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SOVA2 [1]
3 years ago
7

The ice-skating rink charges an hourly fee for skating and $3 to rent skates for the day. Gillian rented skates and skate it for

three hours and was charged $21. which equation represents the cost, c(x), of ice skating as a function of x, The number of hours skating?
A. C(x)=3x+3
B. C(x)=6x+3
C. C(x)=7x+3
D. C(x)=8x+3
Mathematics
1 answer:
djyliett [7]3 years ago
8 0
Subtract $3 from $21 to get the cost of skating for three hours without the rental fee. 21-3=18. $18 was spent on skating for three hours so you'll divide 18 by 3 to get the hourly rate. 18/3=6.
C(x)=6x+3


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16 The perimeter of the pentagon is equal to the perimeter of the square.
STatiana [176]

7x+2

Step-by-step explanation:

add all the sides of the pentagon and make it equal to the perimeter of the square which is 4y

sides of the pentagon are

5x+3

2x+3

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28x+8=4y

7x+2 = y

7x+2 = answer

gathmath

gathmathier6961

another way :

in a standardized test you may want to use real numbers

make x equal a prime number like 3

so then the sides are

18

9

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17

23

=

92

then divide by 4

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8 0
2 years ago
After increase by 30 % it becomes $520?
Art [367]
I believe it is 364.
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4 years ago
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What two numbers multiply to 630 and add to -57
sveta [45]
-15 and -42 both add to -57 and multiply to 630 
5 0
3 years ago
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Factor the trinomial below. x2 – 2x – 35 A. (x – 5)(x + 7) B. (x – 5)(x – 7) C. (x + 5)(x – 7) D. (x + 5)(x + 7)
Sophie [7]
c. (x + 5)(x - 7)    use the FOIL method.

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x² - 2x - 35   

hope that helps, God bless!
3 0
3 years ago
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The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas
Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is = \frac{(n+1)}{2}.

\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5  position orders

10^{th} \ and \ 11^{th} place an average of observations so, the

Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

Q_1 often falls throughout the ordered BELOW average is = \frac{(n+1)}{2} place.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}ordered position

Average 5^{th} \ and \ 6^{th}  location findings

Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5

In  = \frac{(n+1)}{2} the ordered place, Q3 always falls ABOVE the median.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th} ordered At median ABOVE 

Consequently Q_3 falls between 5^{th} \ and \ 6^{th} ABOVE the median position

15^{th}\  and \ 16^{th}place average of observations

\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2}  = 502

\to IQR = Q_3 - Q_1= 502-358.5= 143.5  

\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25

4 0
3 years ago
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